If x=4cos3⁡θ and y=3sin2⁡θ, then 6d2ydx2 at θ=π4 is

Solution:

$\begin{array}{l} x = 4 \cos^{3} θ, y = 3 \sin^{2} θ \Rightarrow \frac{dx}{dθ} = 12 \cos^{2} θ (- \sin θ), \frac{dy}{dθ} = 6 \sin θcos θ \\ \frac{dy}{dx} = \frac{dy / dθ}{dx / dθ} = \frac{6 \sin θcos θ}{- 12 \cos^{2} θsin θ} = - \frac{1}{2 \cos θ} = - \frac{1}{2} \sec θ \\ \frac{d^{2} y}{{dx}^{2}} = - \frac{1}{2} \sec θtan θ \frac{dθ}{dx} = - \frac{1}{2} \frac{\sin θ}{\cos^{2} θ} \times (\frac{- 1}{12 \cos^{2} θsin θ}) = \frac{1}{24 \cos^{4} θ} \\ If θ = \frac{π}{4} than \frac{d^{2} y}{{dx}^{2}} = \frac{1}{24} (\sqrt{2})^{4} = \frac{1}{6} \end{array}$

If $x = 4 \cos^{3} θ and y = 3 \sin^{2} θ, then 6 \frac{d^{2} y}{{dx}^{2}} at θ = \frac{π}{4} is$

Solution:

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